If the equation of a plane P, passing through the intersection of the planes x+4y−z+7=0 and 3x+y+5z=8, is ax+by+6z=15 for some a,b∈R, then the distance of the point (3,2,−1) from the plane P is \_\_\_\_\_\_\_\_\_.
Answer: 3
Solutions · 1
Official solution
Family: (x+4y−z+7)+λ(3x+y+5z−8)=0, i.e. (1+3λ)x+(4+λ)y+(5λ−1)z+(7−8λ)=0. Matching the z- and constant-coefficient ratio to the given form (6 and −15):
\[\frac{5\lambda-1}{6}=\frac{7-8\lambda}{-15}\implies -15(5\lambda-1)=6(7-8\lambda)\implies -27\lambda=27\implies\lambda=-1.\]
Then (1−3,4−1,−15⋅(−1)-scale)=(−2,3,−6), so with scale factor −1: a=2,b=−3, plane 2x−3y+6z−15=0.
\[\text{Distance}=\frac{|2(3)-3(2)+6(-1)-15|}{\sqrt{4+9+36}}=\frac{21}{7}=3.\]